Question #445f7

2 Answers
Feb 14, 2018

5x^2-10x-2=0

Explanation:

If the equation is ax^2+bx+c=0 then alpha+beta= b/a and alpha.beta=c/a.
Comparing ax^2+bx+c=0, x^2+(-2)x+5=0 we get alpha+beta = -2 and alpha.beta=5.
Now, 1/alpha + 1/beta =(alpha+beta)/alpha.beta= -2/5.
Hence new equation is:
x^2+(-2)x+(-2/5)=0
Or,5x^2-10x-2=0

Feb 14, 2018

see a solution process below;

Explanation:

Let A and B be alpha and beta respectively..

x^2-2x + 5 = x^2 -(A+B)x + AB

A + B = 2

AB = 5

When A + B and 1/A+1/B are the roots of the equation,

The expression becomes;

x^2-(A+B+(1/A+1/B)x+[(A+B)(1/A+1/B)]

Therefore;

x^2-(A+B+((A+B)/(AB))x+[(A+B)((A+B)/(AB))]

Substituting the values;

x^2-(2+2/5)x+(2*2/5)

x^2-(12/5)x+4/5

Multiply through by 5

5x^2 - 12x + 4

Hope this helps!